Question

A block of mass *m*_{1} = 1.2 kg initially moving
to the right with a speed of 4.2 m/s on a frictionless, horizontal
track collides with a spring attached to a second block of mass
*m*_{2} = 2.8 kg initially moving to the left with a
speed of

1.0

m/s as shown in figure (a). The spring constant is 535N/m.

What if *m*_{1} is initially moving at 2.6 m/s
while *m*_{2} is initially at rest?(a) Find the
maximum spring compression in this case.

*x* =

Use the condition for maximum compression of the spring and
momentum and energy methods to find the maximum compression.
m

(b) What will be the individual velocities of the two masses
(*v*_{1} and *v*_{2}) after the
spring extended fully again? (That is, when the two masses separate
from each other after the collision is complete.)

v_{1} |
= | m/s to the left |

v_{2} |
= | m/s to the right |

Answer #1

maximum compression occurs when velocities are same

Using conservation of momentum

1.2*2.6 + 0 = (1.2+ 2.8)v

v = 0.78 m/s

combined K.E = 1/2 * 4 * 0.78^{2} = 1.2168 J

Initial kinetic energy = 4.056 J

Therefore,

1/2kx^{2} = 2.8392

x = 0.10302 m

----------------------------------------------------------------------------------

(b) using conservation of momentum

4*0.78 = 1.2*u + 2.8*v

3.12 - 1.2u = 2.8v

v = 3.12 - 1.2u / 2.8

v = (3.12 - 1.2u)^{2} / 7.84

energy is also conserved

1/2*1.2*u^{2} + 1/2*2.8*v^{2} = 4.056

put v^{2} from momentum conservation

1/2*1.2*u^{2} + 1/2*2.8*(3.12 - 1.2u)^{2} / 7.84
= 4.056

0.6u^{2} + 0.1785( 9.7344 + 1.44u^{2} - 7.488u)
= 4.056

0.6u^{2} + 1.7382 + 0.25704u^{2} - 1.3366u =
4.056

0.85704u^{2} - 1.3366u - 2.3178 = 0

solving this quadratic equation will give you the velocity,

Let me know if you have any doubt.

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